Area of a regular decagon! please help! perimeter=240
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Area of a regular decagon! please help! perimeter=240

[From: ] [author: ] [Date: 11-05-27] [Hit: ]
= 1440 sin(72)/sin(18) =4431.86 ~ 4432 sq. u.Note: 1/2(ap) means a =12sin(72)/sin(18)} the height of our triangle or 1/2 width of decagon!......
Ok so I have to find the area of a regular(congruental sided) decagon and I know that the perimeter is 240 and each side is 24. I looked online and found out that the formula is 1/2(ap), so I guess that would make it 1/2(a*240) But idk how to find a. Please help!

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there are 10 isoceles triangles each of base 24 and interior angle 360/10 or 36deg
split the triangle down the middle to form
20 rg. trg orf base=12 and height=12 sin(72deg)/sin(36/2 deg)
where I have used the sine-rule.

Toal area = 20 x area of rg.trg = 20 * 1/2(12){12sin(72)/ sin(18)}
= 1440 sin(72)/sin(18) = 4431.86 ~ 4432 sq. u.

[chk : think circle with peri ~ 240 => r ~ 40 so area ~ pi r^2 = 3 * 40^2 = 3 * 1600 =4800 appro]

Note: 1/2(ap) means a =12sin(72)/sin(18)} the height of our triangle or 1/2 width of decagon!
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